Power Series and: Indeterminates, Probability
1.
Find
sin x
x→0 1 − cos x
sin x
b) lim
+
1 − cos x
x→0
a)
c)
2.
lim
lim
x→0
Note that x now approaches 0 from the positive side only.
sin2 x
1 − cos x
I have denigrated L’Hoptial’s Rule as a way to compute indeterminate forms. I’ve been a bit
unfair; for expressions of the form f (x)/g(x) where x goes to infinity, L’Hopital’s Rule still
applies, but the power series method usually doesn’t work. (Why? Because you can’t usually
write a power series with a = ∞.) However, sometimes power series can still be made to work.
You probably know that, for any power n,
ex
= +∞. This is easy to prove with
x→+∞ xn
lim
L’Hopital’s Rule. Here is a power series proof:
Step 1: for all x > 0, ex >
xn+1
(n+1)!
xn+1 /(n+1)!
= +∞.
x→+∞
xn
ex
Why?
Step 3: lim n = +∞.
x→+∞ x
Step 2:
3.
lim
Why?
Why?
As in class, let pn = the probability that you get your first 6 on the nth toss of a fair die.
∞
X
¡ ¢n−1 1
.
Further,
we
defined
P
(x)
=
pn xn and then found that
We found that pn = 56
6
n=1
x
.
P (x) =
6 − 5x
a)
Let qn = the probability that you don’t get your first 6 on the nth throw, that is, you
either don’t get a 6 at all or you don’t get your first 6. (I assume you keep tossing forever.)
∞
X
x
x
Show that
−
. (This is easy to see if you think about how the
qn xn =
1 − x 6 − 5x
n=1
numbers pn and qn are related.)
b)
In answer to Patrick’s question, I said that P (2) equalled how much you should pay to
play this toss game if you get $2n when your first 6 is on the nth toss. However, show that
∞
X
pn xn doesn’t converge for x = 2. (The expression
what I said can’t be right because
n−=1
x/(6 − 5x) makes fine sense for x = 2, but it doesn’t equal the power series for x = 2.)
What is the interval of convergence for P (x)?
December 5, 1997
Over −→
Power Series and: Indeterminates, Probability
c)
December 5, 1997
Suppose your toss a tetrahedral die instead (4 sides) and you let rn = the probability of
∞
X
getting your first 4 on the nth toss. Find the closed form for R(x) =
rn xn .
n=1
2